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Answer:
We conclude that the true average percentage of organic matter in such soil is different from 3%.
Step-by-step explanation:
We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.
Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.
<em>Let </em><em> = true average percentage of organic matter in such soil</em>
SO, <u>Null Hypothesis</u>, :
= 3% {means that the true average percentage of organic matter in such soil is equal to 3%}
<u>Alternate Hypothesis</u>, :
3% {means that the true average percentage of organic matter in such soil is different than 3%}
The test statistics that will be used here is <u>One-sample t test statistics</u> because we don't know about the population standard deviation;
T.S. = ~
where, = sample mean amount of organic matter = 2.481%
s = sample standard deviation = 1.616%
n = sample of soil specimens = 30
So, <u><em>test statistics</em></u> = ~
= -1.759
<u></u>
<u>Now, P-value of the test statistics is given by;</u>
P-value = P( > -1.759) = <u>0.046</u> or 4.6%
- If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.
- If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.
<em>Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.</em>
Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.